In the given circuit values of I₁, I₂, I₃ are respectively

Correct Answer is : 1.364 A, 6.278 A, 5.91 A1.364\ {A},\; 6.278\ {A},\; 5.91\ {A}1.364 A,6.278 A,5.91 A

1.364\ {A},\; 6.278\ {A},\; 5.91\ {A}

1.97\ {A},\; 3.56\ {A},\; 2.784\ {A}

-0.327\ {A},\; 5.28\ {A},\; 3.197\ {A}

1.523\ {A},\; 4.396\ {A},\; 1.63\ {A}

Correct Answer is : 1.364 A, 6.278 A, 5.91 A1.364\ {A},\; 6.278\ {A},\; 5.91\ {A}1.364 A,6.278 A,5.91 A

Test Index

AP EAPCET 05-Jul-2022 Shift 1 Paper

Section: Physics

Question : 107 of 160

Marks: +1, -0

I_1, I_2, I_3

Solution:

The given circuit diagram is shown as :

Applying KVL in loop

A B D A

, we get

2 I_1+4-I_2\;=0

I_2\;=2 I_1+4

.......(i)

Applying KVL in loop

B C D B

, we get

\;1 (I_1+I_3)-4 (I_2-I_3)-4\;=0

\Rightarrow I_1+I_3-4 I_2+4 I_3-4\;=0

\Rightarrow I_1+5 I_3-4 (2 I_1+4)-4\;=0

\Rightarrow -7 I_1+5 I_3\;=20

\Rightarrow\;I_3\;=\;\frac{20+7 I_1}{5}

.......(ii)

By applying KVL in loop

A D C A

, we get

\;I_2+4 (I_2-I_3)-10\;=0

\Rightarrow 5 I_2-4 I_3\;=10

\Rightarrow\;5 (2 I_1+4) -4 \left(\;\frac{20+7 I_1}{5}\right)\;=10

[From Eqs. (i) and (ii)]

\Rightarrow\; I_1=1.364\ \text{A}

From Eq. (i),

I_2=2 \times 1.364+4=6.278\ \text{A}

From Eq (ii),

I_3=\;\frac{20+7 \times 1.364}{5}=5.91\ \text{A}

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